Let $G = ⟨a⟩$ be a cyclic group of order n. For each integer m, define a map $f_m :G−→G$ by $f_m(x)=x^m$ for every $x∈G$

Question

I have done some parts of this question but other parts I am not sure...

Let $G = ⟨a⟩$ be a cyclic group of order n. For each integer m, define a map $fm :G−→G$ by $f_m(x)=x^m$ for every $x∈G$

(1) $f_m$ is a group homomorphism.

(2) $f_m$ is an automorphism if and only if $gcd(m, n) = 1$

(3) Find the kernel and image of $f_4$ when $n = 10$

The first part I believe is pretty simple. $f_m(xy)=(xy)^m=x^my^m=f_m(x)f_m(y)$.

The next part we have to prove both directions.

For the backwards direction I think I am more clear with what I have to do. We know from the first part $fm$ is a homomorphism. So to prove it to be a automorphism we must show $fm$ is one to one and onto. Let x and y exist in G and |G|=u. Then $x^u=e=y^u$. Since the gcd(m,n)=1 by defintion there exists a u and v in the integers s.t. $mu+nv=1$. We can rearrange this and see $mu=1-nv$.

First we will show this is 1-1:

$x^m=y^m$

$x^{mu}=y^{mu}$

$x^{1-nv}=y^{1-nv}$

$x*x^{n(-v)}=y*y^{n(-v)}$

xe=ye

x=y

Now we will prove it is onto. We need to pick a g that exists in G and find x that exists in G s.t. $fm(x)=x^m=y$. y=$y^1$=$y^{mu+nv}$=$y^{mu}y^{nv}=(y^{u})^m(y^{n})^v$. I'm not sure what to do here after this...

The foward direction seems to be much harder. We are assuming $fm$ is an isomorphism and proving the gcd(m,n)=2. $fm$ is 1-1 and we can assume d=(m,n)>1. I believe for this we can use the formula $ord(x)=n/gcd(m,n)=d>1$ but I am not sure how to get here. For the beginning of the foward direction do we let x exist in G s.t $fm(x)=e$ or is that incorrect?

The third part I really have no idea. I know the ker here is when $fm(x)=e$

Answer 1

Hint

(2) For the ontoness, note that $x^n=e$ since your group has $n$ elements and you want to find an $x$ such that $x^m=y$.

Then $$x=x^{mu+nv}=y^u$$

Note Since $f: G \to G$ is one-to-one and $G$ is finite, you can deduce directly that $f$ is onto, this is a well known set theory fact.

Show that this $y$ works.

For the direct implication: Assume by contradiction that $gcd(m,n)=d >1$. Write $m=dm',n=dn'$.

What is $f(a^{n'})$?

(3) $G=\{ e, a, a^2,..., a^9 \}$ and $a^{10}=e$.

By definition $$\ker(f_4)=\{ a^k : f_4(a^k)=e \}$$

This is equivalent to $a^{4k}=e$, which only happens if $10 |4k$.

Also $$Im(f)=\{ f_4(a^k) : 0 \leq k \leq 9 \} = \{e, a^4, a^8,..., a^{36} \}$$

Use $a^{10}=e$ to write $Im(f)$ explicitely.